Is the function given below continuous/differentiable at $x=3$ ? $f(x)=\begin{cases} x^2&,&x<3 \\\\ 6x-9&,&x\geq3 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Answer: Checking for continuity at $x=3$ For the function to be continuous at $x=3$, we need the two-sided limit $\lim_{x\to 3}f(x)$ to exist and be equal to $f(3)$. This is the same as requiring that the two one-sided limits $\lim_{x\to 3^-}f(x)$ and $\lim_{x\to 3^+}f(x)$ exist and are equal to $f(3)$. According to $f$ 's definition, $f(3)=6\cdot3-9=9$. $\lim_{x\to 3^-}f(x)$ $x^2$ evaluated at $x=3$ is equal to $9$. Since $x^2$ is continuous, we can be certain that $\lim_{x\to 3^-}f(x)=9$. $\lim_{x\to 3^+}f(x)$ $6x-9$ evaluated at $x=3$ is equal to $9$. Since $6x-9$ is continuous, we can be certain that $\lim_{x\to 3^+}f(x)=9$. We saw that the two one-sided limits exist and are equal to $f(3)$, so the function is continuous at $x=3$. Checking for differentiability at $x=3$ For the function to be differentiable at $x=3$, we need the two-sided limit $\lim_{x\to 3}\dfrac{f(x)-f(3)}{x-3}=\,\lim_{x\to 3}\dfrac{f(x)-9}{x-3}$ to exist. This is the same as requiring that the two one-sided limits $\lim_{x\to 3^-}\dfrac{f(x)-9}{x-3}$ and $\lim_{x\to 3^+}\dfrac{f(x)-9}{x-3}$ exist and have the same value. $\lim_{x\to 3^-}\dfrac{f(x)-9}{x-3}=6$ $\lim_{x\to 3^+}\dfrac{f(x)-9}{x-3}=6$ We saw that the two one-sided limits exist and are equal, so the function is differentiable at $x=3$. In conclusion, the function is both continuous and differentiable at $x=3$.